Exercises — Simple harmonic motion — definition, restoring force F = −kx

SHM ke liye ek aur sirf ek test hai: kya force seedha ghar ki taraf push-back hai? Symbols mein, $F=-kx$ ke saath ek positive constant $k$ aur ek minus sign.

• (a) $F=-5x$: $x$ mein linear hai, minus sign present hai → SHM, $k=5\text{N/m}$ ke saath.
• (b) $F=-5x^3$: yeh ghar ki taraf zaroor point karta hai (jab $x>0$ tab negative), toh oscillate karta hai — lekin strength $x^3$ ki tarah badhti hai, $x$ ke proportional nahi. Toh yeh SHM nahi hai. Iska period amplitude par depend karta.
• (c) $F=+5x$: sign plus hai, toh force ghar se door push karta hai. Yeh bhaag jaata hai, kabhi oscillate nahi karta → SHM nahi (yeh ek unstable, "anti-restoring" force hai).

Neeche ki figure teen force laws plot karti hai taki tum farq dekh sako: sirf negative slope wali straight line jo origin se jaati hai (blue wali) SHM hai.

Answer: sirf (a).

SHM ki definition hai $a=-{\omega }^{2}x$. Toh $x$ ko multiply karne wala number hi ${\omega }^2$ hai. ${\omega }^2=64\Rightarrow \omega =\sqrt{64}=8\text{rad/s}.$ Period ghadi ka ek pura chakkar hai, $2\pi$ radians, $8$ rad/s par: $T=\frac{2\pi }{\omega }=\frac{2\pi }{8}=\frac{\pi }{4}\approx 0.785\text{s}.$

Sawaal: "iska clock kitni tez tick karta hai?" → tool hai $\omega =\sqrt{k/m}$. $\omega =\sqrt{\frac{80}{0.20}}=\sqrt{400}=20\text{rad/s}.$ Period (seconds per cycle): $T=\frac{2\pi }{\omega }=\frac{2\pi }{20}=0.3142\text{s}.$ Frequency (cycles per second) bas reciprocal hai: $f=\frac{1}{T}=\frac{20}{2\pi }=3.183\text{Hz}.$

Sawaal: "diye gaye position par, time solve kiye bina, kitni speed?" → energy tool $v=\omega \sqrt{A^2-x^2}$. $v=20\sqrt{0.05^2-0.03^2}=20\sqrt{0.0025-0.0009}=20\sqrt{0.0016}=20(0.04)=0.80\text{m/s}.$ Max speed ghar par hoti hai ($x=0$), jahan saari energy motion mein hoti hai: $v_{\max }=\omega A=20(0.05)=1.0\text{m/s}.$

SHM mein acceleration hai $a=-{\omega }^{2}x$, size mein sabse badi jab $|x|$ sabse bada ho — yaani turning points $x=\pm A$ par. $a_{\max }={\omega }^{2}A=(20{)}^2(0.05)=400(0.05)=20{\text{m/s}}^2,$ yeh extremes $x=\pm A$ par hota hai, jahan block ek pal ke liye ruka hua hota hai lekin restoring force sabse strong hoti hai.

Total budget fixed hai: $E=\frac{1}{2}k{A}^2$, bata hua: $E=\underset{KE}{\underbrace{\frac{1}{2}m{v}^2}}+\underset{PE}{\underbrace{\frac{1}{2}k{x}^2}}$. Hum chahte hain $KE=PE$, matlab har ek aadha budget rakhta hai: $\frac{1}{2}k{x}^2=\frac{1}{2}E=\frac{1}{2}\left(\frac{1}{2}k{A}^2\right).$ Ab $x$ isolate karte hain. $\frac{1}{2}k$ kyun cancel karein? Dono taraf "$\frac{1}{2}k$ times length-squared" hai, toh common factor $\frac{1}{2}k$ se dono taraf divide karne par sirf lengths ke baare mein ek clean statement reh jaata hai — stiffness $k$ kabhi yeh affect nahi karta ki split kahaan hoti hai, toh hum ise hata dete hain: $x^2=\frac{1}{2}{A}^2.$ Square root kyun lein? Hum $x^2$ jaante hain lekin $x$ khud chahiye, aur square root woh operation hai jo squaring ko undo karta hai. Kyunki displacement ghar ke kisi bhi taraf ho sakta hai, dono signs physical hain: $x=\pm \frac{A}{\sqrt{2}}\approx \pm 0.707A.$ Answer: $x=A/\sqrt{2}$ par (ghar ke kisi bhi taraf) energy fifty-fifty split hoti hai. Neeche ki figure dono energies alag alag position ke against plot karti hai: pink potential curve $U=\frac{1}{2}k{x}^2$ valley ki tarah upar uthti hai, blue kinetic curve $KE=E-U$ ulti taraf girti hai, aur yellow dashed line unka constant total $E$ hai. Woh exactly $x=A/\sqrt{2}$ par cross karti hain — crossing point wahan hai jahan har ek aadha total rakhta hai.

Do springs in parallel dono mass ko kheenchte hain; unki pushes add hoti hain. Agar har ek $-kx$ deta hai, saath mein woh $-2kx$ dete hain, toh effective stiffness hai $k_{\text{eff}}=2k$. $T_A=2\pi \sqrt{\frac{m}{k}},T_B=2\pi \sqrt{\frac{m}{2k}}=\frac{T_A}{\sqrt{2}}.$ Answer: parallel pair stiffer hai, toh B tez oscillate karta hai: $T_B=T_A/\sqrt{2}\approx 0.707T_A$.

Standard form $x=A\cos (\omega t+\varphi )$ se compare karo. Yahan $\varphi$ phase constant hai — cosine mein baka hua ek fixed head-start angle jo kehta hai ki clock-start $t=0$ par motion apne cycle mein kahaan thi*. Yeh time ke saath nahi badlata; bas poori wave ko sideways shift karta hai. Is problem mein $\varphi =\pi /3$ hai.

• (a) Amplitude $A=0.10\text{m}$ (aage wala number).
• (b) $\omega =4\text{rad/s}$ ($t$ ko multiply karne wala number), toh $T=2\pi /4=1.571\text{s}$.
• (c) $t=0$ par cosine ke andar ka angle bas $\varphi =\pi /3$ hai: $x(0)=0.10\cos (\pi /3)=0.10(0.5)=0.05\text{m}.$ Velocity position ka time-derivative hai, $v=-A\omega \sin (\omega t+\varphi )$ ($\cos$ differentiate karne par $-\sin$ aata hai, aur chain rule $\omega$ neeche laata hai): $v(0)=-0.10(4)\sin (\pi /3)=-0.4(0.8660)=-0.3464\text{m/s}.$ Minus sign kehta hai ki $t=0$ par yeh ghar ki taraf wapas ja raha hai.

Us tool se shuru karo jo $T$ aur $k$ link karta hai: $T=2\pi \sqrt{m/k}$. Hum $T$ aur $m$ jaante hain, hum $k$ chahte hain — toh hume $k$ ko square root ke andar se unrap karna hoga. Pehle dono sides square kyun karein? $k$ ek square root ke neeche fansa hua hai; squaring woh operation hai jo square root ko hata deta hai, toh yeh $k$ ko radical se azad karta hai: $T^2=4{\pi }^2\frac{m}{k}.$ $k$ se multiply aur $T^2$ se divide kyun karein? $k$ abhi denominator mein hai; dono sides ko $k$ se multiply karne par woh upar aa jaata hai, aur $T^2$ se divide karne par $k$ ek taraf akela reh jaata hai: $k=\frac{4{\pi }^{2}m}{T^2}.$ Ab numbers substitute karo: $k=\frac{4{\pi }^2(0.30)}{(0.50{)}^2}=\frac{4(9.8696)(0.30)}{0.25}=\frac{11.844}{0.25}=47.37\text{N/m}.$ Answer: $k\approx 47.4\text{N/m}$.

Uniform circular motion ki shadow exactly SHM hai: shadow ka displacement hai $x=R\cos (\omega t)$. Toh wheel ki radius amplitude ban jaati hai, aur wheel ki spin rate SHM $\omega$ ban jaati hai.

• Amplitude: $A=R=0.20\text{m}$.
• Max speed = peg ki actual rim speed $=\omega R=6(0.20)=1.2\text{m/s}$.
• Max acceleration = peg ka centripetal acceleration $={\omega }^{2}R=36(0.20)=7.2{\text{m/s}}^2$. Answer: $A=0.20\text{m}$, $v_{\max }=1.2\text{m/s}$, $a_{\max }=7.2{\text{m/s}}^2$. Figure mein pink peg circle par sawaar hai jabki yellow shadow neeche wall par slide karta hai; shadow ki swing-width exactly circle ka radius hai, aur yeh middle cross karte waqt sabse tez hota hai.

(a) Turning point par saari energy spring mein hoti hai: $E=\frac{1}{2}k{A}^2$. $A$ ke liye invert kyun karein? Hum energy aur stiffness jaante hain aur reach chahiye, toh hum $A$ unrap karte hain: $2$ se multiply karo, $k$ se divide karo, root lo (root squaring ko undo karta hai): $A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.30)}{60}}=\sqrt{0.01}=0.10\text{m}.$ (b) Ghar par saari energy kinetic hai: $E=\frac{1}{2}m{v}_{\max }^2$, toh same unwrapping se ($2$ se multiply, $m$ se divide, root): $v_{\max }=\sqrt{\frac{2E}{m}}=\sqrt{\frac{0.60}{0.40}}=\sqrt{1.5}=1.2247\text{m/s}.$ (c) $x=A/2=0.05\text{m}$ par, $v=\omega \sqrt{A^2-x^2}$ use karo $\omega =\sqrt{k/m}=\sqrt{150}=12.247\text{rad/s}$ ke saath: $v=12.247\sqrt{0.10^2-0.05^2}=12.247\sqrt{0.0075}=12.247(0.08660)=1.0607\text{m/s}.$ Answers: $A=0.10\text{m}$, $v_{\max }=1.225\text{m/s}$, $v(A/2)=1.061\text{m/s}$.

Trial solution ko Newton's law mein substitute karo. Pehle hume uska second time-derivative chahiye. Ek baar differentiate karne par, $\dot{x}=-A\omega \sin (\omega t+\varphi )$; dobara differentiate karne par, $\ddot{x}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x.$ Ise $m\ddot{x}=-kx$ mein daalo: $m(-{\omega }^{2}x)=-kx.$ Coefficients match kyun kar sakte hain? Yeh equation har pal true honi chahiye, aur $x(t)$ kai alag alag values se guzarta hai jab block move karta hai. $-m{\omega }^{2}x=-kx$ un saari alag alag values of $x$ ke liye simultaneously true hone ka ek hi tarika hai ki $x$ ko multiply karne wale constants equal hon. Isliye: $m{\omega }^2=k\Rightarrow \omega =\sqrt{\frac{k}{m}},T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}.$ Result dekho: amplitude $A$ bilkul cancel ho gaya — yeh $m\ddot{x}=-kx$ ke dono sides par tha ($x$ ke andar aur $\ddot{x}$ ke andar) aur divide ho gaya, toh yeh kabhi $m$, $k$, aur $\omega$ ke beech ke relation ko nahi chhuta. Isliye $T$ kabhi $A$ par depend nahi kar sakta.

Physical reason ki intuition kyun fail karti hai: $A$ double karne par block ko travel karne ki distance double ho jaati hai, toh naively tumhe longer trip ki ummeed hoti hai. Lekin $A$ double karne par har corresponding point par displacement bhi double ho jaata hai, aur kyunki $F=-kx$ linear hai, yeh restoring force aur isliye top speed bhi double kar deta hai ($v_{\max }=\omega A$). Lamba raasta, proportionally tez yatra — dono effects exactly cancel hote hain. Yeh exact cancellation sirf isliye hoti hai kyunki $F$ strictly $x$ ke proportional hai; $F=-k{x}^3$ jaisi nonlinear law ke liye cancellation imperfect hogi aur period zaroor amplitude par depend karta.

• (a) $A=0$: block exactly ghar par baitha hai $x(t)=0$ saare $t$ ke liye. Har quantity — speed, force, energy — zero hai. Yeh trivial equilibrium hai; technically yeh zero amplitude ke saath SHM hai lekin kuch move nahi karta.
• (b) $m\to \infty$: $\omega =\sqrt{k/m}\to 0$, toh $T=2\pi \sqrt{m/k}\to \infty$. Bahut bhaari mass itna sluggish hai ki ek oscillation mein hamesha lag jaata hai — practically hilta hi nahi. Limit hai ek stationary block.
• (c) $k\to 0$: spring limp ho jaati hai, koi restoring force nahi. $\omega \to 0$, $T\to \infty$. Koi push-back nahi hone par, nudged block bas drift karta rehta hai constant velocity se hamesha — koi oscillation nahi. Yeh woh boundary hai jahan SHM exist karna band kar deta hai: restoring force ke liye $k>0$ chahiye.

Yeh limits formula ki honesty confirm karte hain: yeh exactly wahan "no oscillation" predict karta hai jab physics apni restoring force ya accelerate karne ki ability kho deta hai.

$x=0$ par minimum ke paas, potential ko power series (Taylor series — "kisi bhi smooth curve ko polynomial se approximate karo" ka tool) ke roop mein expand karo: $U(x)\approx U(0)+U^{\prime }(0)x+\frac{1}{2}{U}^{\prime \prime }(0)x^2+\cdots$ Ab do facts use karo jo isliye true hain kyunki $x=0$ stable minimum hai:

1. Valley ke bottom par slope flat hota hai, toh $U^{\prime }(0)=0$. Yeh linear term ko khatam kar deta hai.
2. Valley upar ki taraf bend karti hai, toh curvature $U^{\prime \prime }(0)>0$ hai.

Bacha hua constant $U(0)$ bas ek baseline hai — ek fixed height jo hum zero set kar sakte hain, kyunki sirf energy mein changes force produce karte hain (force negative slope hai, aur constant ka slope zero hota hai). Khatam hua linear term aur inert constant drop karne par: $U(x)\approx \frac{1}{2}{U}^{\prime \prime }(0)x^2.$ Yeh ek parabola hai — exactly spring potential $\frac{1}{2}k{x}^2$ ki shape. Ab force nikalo energy ka negative slope lekar (force energy ka minus derivative hai): $F=-\frac{dU}{dx}\approx -U^{\prime \prime }(0)x=-k_{\text{eff}}x,\boxed{k_{\text{eff}}=U^{\prime \prime }(0)}.$ Conclusion: har stable equilibrium small nudges ke liye spring ki tarah behave karta hai, effective stiffness bottom par energy valley ki curvature ke equal hoti hai, $k_{\text{eff}}=U^{\prime \prime }(0)$. Yahi ek fact hai kyun SHM har jagah milta hai — pendulums, atoms in crystals, vibrating molecules — sab parabolic valley bottom share karte hain.

Level 1 — SHM ke liye ek test kya hai? ::: Force $F=-kx$ honi chahiye: $x$ mein linear, minus sign ke saath aur $k>0$. Level 2 — $\omega =20$, $A=0.05$ ke liye $x=0.03$ par speed? ::: $v=20\sqrt{0.05^2-0.03^2}=0.80\text{m/s}$. Level 3 — $KE=PE$ kahaan hota hai? ::: $x=A/\sqrt{2}\approx 0.707A$ par. Level 4 — $T=0.5\text{s}$, $m=0.3\text{kg}$ ke liye $k$ design karo? ::: $k=4{\pi }^{2}m/T^2\approx 47.4\text{N/m}$. Level 5 — General potential se $k_{\text{eff}}$ kya hai? ::: $k_{\text{eff}}=U^{\prime \prime }(0)$, uske minimum par energy valley ki curvature.